package StackPractice;

public class IsLandNumDFS {
    /**
     * 岛屿数量
     * 给你一个由'1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。
     * <p>
     * 岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
     * <p>
     * 此外，你可以假设该网格的四条边均被水包围。
     * <p>
     * 示例 1：
     * <p>
     * 输入：grid = [
     * ["1","1","1","1","0"],
     * ["1","1","0","1","0"],
     * ["1","1","0","0","0"],
     * ["0","0","0","0","0"]
     * ]
     * 输出：1
     * 示例 2：
     * <p>
     * 输入：grid = [
     * ["1","1","0","0","0"],
     * ["1","1","0","0","0"],
     * ["0","0","1","0","0"],
     * ["0","0","0","1","1"]
     * ]
     * 输出：3
     */

    public int numIslands(char[][] grid) {
        int col = grid.length;
        int row = grid[0].length;
        int numIsLand = 0;
        for (int i = 0; i < col; i++) {
            for (int j = 0; j < row; j++) {
                if (grid[i][j] == '1') {
                    numIsLand++;
                    dfs(grid, i, j);
                }
            }
        }
        return numIsLand;
    }

    //深度优先搜索
    private void dfs(char[][] grid, int i, int j) {
        int col = grid.length;
        int row = grid[0].length;
        if (i < 0 || j < 0 || i + 1 > col || j+1 > row || grid[i][j] == '0') {
            return;
        }
        grid[i][j] = '0';
        dfs(grid, i + 1, j);
        dfs(grid, i - 1, j);
        dfs(grid, i, j + 1);
        dfs(grid, i, j - 1);
    }
}
